2. Classifications of Mixtures Heterogeneous Mixtures—composed of different types of phases of substances - ex: Fruit salad Granite Homogeneous Mixtures—the same throughout (substances have dissolved in one another) - ex: Salt water Alloys (metal mixtures)

3. Types of Mixtures Solutions Suspensions Colloids

4. Solutions are: A homogeneous mixture of two or more substances in a single phase Composed of: 1.Solvent- the substance that does the dissolving 2.Solute- the substance that is being dissolved Example: In sugar water, water is the solvent and sugar is the solute.

5. Suspensions A mixture in which the particles are so large that they settle out unless the mixture is constantly stirred or agitated. Ex. A jar of muddy water Ex: Orange juice

6. Colloids A mixture in which the particles are intermediate in size between those in solutions and suspensions. Particles are suspended in mixture Ex. Jello, Paints, Milk, Mayonnaise, Fog, Cheese

7. Tyndall Effect Colloids appear to be homogeneous mixtures (aka solutions) because the individual particles cannot be seen…however, they are not true solutions. The particles are large enough to scatter light. This effect is known as the Tyndall Effect.

8. Types of Solutions: ***Type of solution is determined by the solvent: Gaseous-mixture of two or more gases Liquid- solvent is a liquid Solid- solvent is a solid Examples of: 1.Gaseous- air, scuba tanks 2.Liquid- tea, Kool- aid, cokes, salt water 3.Solid- metal alloys, amalgams (dental fillings)

9. Some solutions conduct electricity : Electrolytes- a solution that conducts electricity as a result of the formation of ions in solution (examples: salt water, vinegar) Nonelectrolyte- a solution that does not conduct electricity because there is no formation of ions in solution (example: sugar in oil)

10. What solutions of electrolytes look like versus solutions that are not electrolytes: Electrolyte Nonelectrolyte

11. Factors that Affect Rates of Dissolving Solids in Liquids: 1.Surface Area- increasing surface area increases the rate of dissolving 2.Agitation- stirring or shaking increases rate of dissolving 3.Heat- heating the solvent will increase the rate of dissolving

12. Solubility The amount of a substance that is dissolved at solution equilibrium in a specific amount of solvent at a specified temperature. Factors Affecting Solubility: 1.Nature of Solute and Solvent- “like dissolves like”- polar dissolves polar-nonpolar dissolves nonpolar 2.Pressure- changes in pressure have little affect on dissolving solids in liquids but an increase in pressure will increase the solubility of gases in liquids 3.Temperature a)Increasing temp., decreases gas solubility b)Increasing temp., increases solid solubility

13. 3 Classifications of Solutions Saturated- a solution that contains the max. amount of dissolved solute at a given temp. Unsaturated- a solution that contains less solute than a saturated solution at a given temp. Supersaturated- a solution that contains more dissolved solute than a saturated solution at a given temp.

14. Solubility Curves: Show how much solute can go into solution with a given amount of solvent at different temperatures.

15. Solubility Curve:

16. Colligative Properties A property that depends on the number of solute particles but is independent of their nature 3 Colligative Properties 1.Vapor Pressure Lowering 2.Freezing Point Depression 3.Boiling Point Elevation

17. Vapor Pressure Lowering- the vapor pressure of a solvent containing a nonvolatile solute is lower than the vapor pressure of the pure solvent at any temp. Freezing Point Depression- the freezing point of a solvent containing a solute will be lower than the pure solvent Boiling Point Elevation- the boiling point of a solvent containing a solute will be higher than the pure solvent

18. Solution Concentration: Concentration is a measurement of the amount of solute in a given amount of solvent or solution Can be expressed qualitatively (using words) or quantitatively (using numbers) Qualitative Terms: 1.Dilute- relatively small amount of solute compared to solvent 2. Concentrated- relatively large amount of solute in a solvent

19. Quantitative Terms: Percent by Mass Molarity Molality

20. Percent by Mass The number of grams of solute dissolved in 100 g of water Percent by mass = mass solute x100 (mass solute + mass solvent)

21. Examples: A solution of sodium chloride is prepared by dissolving 5 g of salt in 550 g of water. What is the concentration as given by percent by mass? Percent by mass = mass solute x100 (mass solute + mass solvent) Percent by mass = __5__ x 100 = 0.9% (5 + 550) Answer: 0.9%

22. Examples: What is the percent by mass of a solution prepared by dissolving 4 g of acetic acid in 35 g of water. Percent by mass = mass solute x100 (mass solute + mass solvent) Percent by mass = __4__ x 100 = 10.26% (4 + 35) Answer: 10.26%

23. Molarity Symbolized by M Units of mol L Describes how many moles of solute are present per liter of solution

24. Examples: What is the molarity of 3.5 L of solution that contains 90 g of sodium chloride? Molarity = mol L 90 g NaCl x 1 mol NaCl = 1.54 mol NaCl 58.44 g NaCl Molarity = mol = 1.54 mol solute = 0.44 M L 3.5 L solution Answer: 0.44 M

25. Examples: How many moles of HCl are present in 0.8 L of a 0.5 M HCl solution? Molarity = 0.5 M Molarity = mol L 0.5 = mol HCl 0.8 Answer: mol HCl = 0.4 mol

26. Examples: How many grams of sodium chloride will be required to make 555 mL of a 1.45 M solution? Molarity = 1.45 M 555 mL = 0.555 L Molarity = mol L 1.45 = mol NaCl 0.555 mol NaCl = 0.80475 mol 0.80475 mol NaCl x 58.44 g 1 mol NaCl  47.03 g NaCl

27. More Examples: How many liters of solution can be prepared if 78.9 g of sodium chloride is used to make a 3.00 M solution? 78.9 g NaCl x 1 mol NaCl = 1.35 mol NaCl 58.44 g NaCl Molarity = mol L 3 = 1.35 L Answer: 0.45 L

28. More Examples: What is the molarity of a solution that is prepared by using 20 g of sodium hydroxide in enough water to make a 2 L solution? 20 g NaOH x 1 mol NaOH = 0.5 mol NaOH 39.997 g NaOH Molarity = mol = 0.5 mol = 0.25 M L 2 L Answer: 0.25 M

29. Dilution Problems Use the equation: M 1 V 1 = M 2 V 2 Where M 1 = molarity 1 V 1 = volume 1 M 2 = molarity 2 V 2 = volume 2

30. Examples: What is the molarity of a solution that is made by diluting 50 mL of a 4.74 M solution to 250 mL? M 1 V 1 = M 2 V 2 (4.74)(0.05L) = (M 2 )(0.25) M 2 = 0.948 M Answ: 0.948 M

31. Molality Symbolized by m Units are mol solute kg of solvent Describes how many moles of solute are present per kg of solvent.

32. Examples: A solution contains 17.1 g of sucrose (C 12 H 22 O 11 ) dissolved in 125 g of water. Find the molal concentration. Molality = mol solute kg solvent Solute: 17.1 g C 12 H 22 O 11 x 1 mol C 12 H 22 O 11 = 0.05 mol C 12 H 22 O 11 342.3 g C 12 H 22 O 11 Solvent: 125 g water = 0.125 kg Molality = mol solute = 0.05 = 0.400 m kg solvent 0.125 Answer: 0.400 m

33. Examples: How much iodine (in grams) must be added to prepare a 0.480 m solution of iodine in carbon tetrachloride (CCl 4 ) if 100 g of CCl 4 is used? Molality = mol solute kg solvent Solvent: 100 g CCl 4 = 0.100 kg CCl 4 Molality = mol solute kg solvent 0.480 = mol solute 0.100 kg Mol solute = 0.048 mol iodine 0.048 mol I 2 x 253.8 g I 2 = 12.2 g 1 mol I 2 Answer: mol solute = 12.2 g

34. More Examples: What is the molality of a solution composed of 2.55 g of acetone (CH 3 ) 2 CO, dissolved in 200 g of water? Molality = mol solute kg solvent Solute: 2.55 g (CH 3 ) 2 CO x 1 mol (CH 3 ) 2 CO = 0.0439 mol (CH 3 ) 2 CO 58.08 g (CH 3 ) 2 CO Solvent: 200 g water = 0.200 kg water Molality = mol solute kg solvent Molality = 0.0439 / 0.200 = 0.220 m Answer: 0.220 m

35. More Examples: What quantity, in grams, of methanol (CH 3 OH) is required to prepare a 0.244 m solution in 400 g of water? Molality = mol solute kg solvent Solvent: 400 g water = 0.400 kg 0.244 = mol solute 0.400 kg mol solute = 0.0976 kg CH 3 OH 0.0976 mol CH 3 OH x 32.04 g CH 3 OH = 3.13 g 1 mol CH 3 OH Answer: 3.13 g

36. How many grams of AgNO 3 are needed to prepare a 0.125 m solution in 250 mL of water? (1 mL of water = 1g and 1 L of water = 1 kg) Molality = mol solute kg solvent Solvent: 250 mL water = 0.250 L water 0.125 = mol solute 0.250 kg mol solute = 0.03125 mol AgNO 3 0.03125 mol AgNO 3 x 169.87 g AgNO 3 = 5.31 g AgNO 3 1 mol AgNO3 Answ: 5.31 g AgNO 3

38. Examples: 2H 3 PO 4 + 3Ca(OH) 2  Ca 3 (PO 4 ) 2 +6 H 2 O If 2 L of 4 M phosphoric acid is used, how many grams of water could be formed? Molarity = mol solute 4 = mol phosphoric acid mol phos acid = 8 mol L solution 2 liters 8 mol H 3 PO 4 6 mol H 2 0 18.01532 g H 2 O = 432.368 g H 2 O 2 mol H 3 PO 4 1 mol H 2 O

39. Example 2H 3 PO 4 + 3Ca(OH) 2  Ca 3 (PO 4 ) 2 +6 H 2 O If 6 g of calcium hydroxide is used, how many liters of a 5 M acid solution would be needed? Molarity = mol solute L solution 5 M = mol H 3 PO 4 L solution **Must solve for mol H 3 PO 4 so we can plug it into the equation above 6 g Ca(OH) 2 1 mol Ca(OH) 2 2 mol H 3 PO 4 = 0.054 mol H 3 PO 4 74.0932 g Ca(OH) 2 3 mol Ca(OH) 2 5 M = mol H 3 PO 4 5 = 0.054 mol H 3 PO 4 L solution = 0.011 liters L solution L solution

40. 2H 3 PO 4 + 3Ca(OH) 2  Ca 3 (PO 4 ) 2 +6 H 2 O When 200 mL of 6 M phosphoric acid is used, how many mL of a 3 M calcium hydroxide solution would be required? Molarity = mol solute L solution 6 = mol phosphoric acid 0.200 L mol phosphoric acid = 1.2 mol H 3 PO 4 Molarity = mol solute L solution 3 = mol CaOH L solution